Question

Who can solve this geometry question​

Answer 1

From the fig.,

$\longrightarrow\sf{BC=5}$

$\longrightarrow\sf{x+y+2r=5\quad\quad\dots(1)}$

In a triangle PQR, if ∠P = ∠Q or PR = QR = a, then,

$\longrightarrow\sf{PQ=2a\sin\left(\dfrac{\angle R}{2}\right)}$

It's a case of law of cosines.

Then, from ΔCDE,

$\longrightarrow\sf{DE=2x\sin\left(\dfrac{\alpha}{2}\right)\quad\quad\dots(2)}$

From ΔDME, since the quadrilateral CDME is cyclic,

• $\sf{\angle DME=180^o-\alpha}$

Then,

$\longrightarrow\sf{DE=2r\sin\left(\dfrac{180^o-\alpha}{2}\right)}$

$\longrightarrow\sf{DE=2r\sin\left(90^o-\dfrac{\alpha}{2}\right)}$

$\longrightarrow\sf{DE=2r\cos\left(\dfrac{\alpha}{2}\right)\quad\quad\dots(3)}$

From (2) and (3),

$\longrightarrow\sf{2x\sin\left(\dfrac{\alpha}{2}\right)=2r\cos\left(\dfrac{\alpha}{2}\right)}$

$\longrightarrow\sf{x=\dfrac{r}{\tan\left(\dfrac{\alpha}{2}\right)}\quad\quad\dots(4)}$

Similarly, on considering triangles GNF and GBF, since the quadrilateral GNFB is also cyclic, we get,

$\longrightarrow\sf{y=\dfrac{r}{\tan\left(\dfrac{\beta}{2}\right)}\quad\quad\dots(5)}$

From ΔABC, we have,

• $\sf{\cos\alpha=\dfrac{3}{5}}$

• $\sf{\cos\beta=\dfrac{4}{5}}$

We know that,

$\longrightarrow\sf{\tan\left(\dfrac{\theta}{2}\right)=\sqrt{\dfrac{1-\cos\theta}{1+\cos\theta}}}$

Hence,

$\longrightarrow\sf{\tan\left(\dfrac{\alpha}{2}\right)=\sqrt{\dfrac{1-\cos\alpha}{1+\cos\alpha}}}$

$\longrightarrow\sf{\tan\left(\dfrac{\alpha}{2}\right)=\sqrt{\dfrac{1-\dfrac{3}{5}}{1+\dfrac{3}{5}}}}$

$\longrightarrow\sf{\tan\left(\dfrac{\alpha}{2}\right)=\dfrac{1}{2}}$

And,

$\longrightarrow\sf{\tan\left(\dfrac{\beta}{2}\right)=\sqrt{\dfrac{1-\cos\beta}{1+\cos\beta}}}$

$\longrightarrow\sf{\tan\left(\dfrac{\beta}{2}\right)=\sqrt{\dfrac{1-\dfrac{4}{5}}{1+\dfrac{4}{5}}}}$

$\longrightarrow\sf{\tan\left(\dfrac{\beta}{2}\right)=\dfrac{1}{3}}$

Hence (4) becomes,

$\longrightarrow\sf{x=\dfrac{r}{\left(\dfrac{1}{2}\right)}}$

$\longrightarrow\sf{x=2r}$

And (5) becomes,

$\longrightarrow\sf{y=\dfrac{r}{\left(\dfrac{1}{3}\right)}}$

$\longrightarrow\sf{y=3r}$

Finally, (1) becomes,

$\longrightarrow\sf{2r+3r+2r=5}$

$\longrightarrow\sf{7r=5}$

$\longrightarrow\sf{\underline{\underline{r=\dfrac{5}{7}}}}$

Answer 2

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HERE IS UR ANSWER

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$From the fig.,</p><p>\longrightarrow\sf{BC=5}⟶BC=5\\</p><p>\longrightarrow\sf{x+y+2r=5\quad\quad\dots(1)}⟶x+y+2r=5…(1)\\</p><p>In a triangle PQR, if ∠P = ∠Q or PR = QR = a, then,\\</p><p>\longrightarrow\sf{PQ=2a\sin\left(\dfrac{\angle R}{2}\right)}\\⟶PQ=2asin(2∠R)\\</p><p>It's a case of law of cosines.\\</p><p>Then, from ΔCDE,\\</p><p>\longrightarrow\sf{DE=2x\sin\left(\dfrac{\alpha}{2}\right)\quad\quad\dots(2)}⟶DE=2xsin(2α)…(2)\\</p><p>From ΔDME, since the quadrilateral CDME is cyclic,\\</p><p>\sf{\angle DME=180^o-\alpha}∠DME=180o−α\\</p><p>Then,\\</p><p>\longrightarrow\sf{DE=2r\sin\left(\dfrac{180^o-\alpha}{2}\right)}⟶DE=2rsin(2180o−α)\\</p><p>\longrightarrow\sf{DE=2r\sin\left(90^o-\dfrac{\alpha}{2}\right)}⟶DE=2rsin(90o−2α)\\</p><p>\longrightarrow\sf{DE=2r\cos\left(\dfrac{\alpha}{2}\right)\quad\quad\dots(3)}⟶DE=2rcos(2α)…(3)\\</p><p>From (2) and (3),\\</p><p>\longrightarrow\sf{2x\sin\left(\dfrac{\alpha}{2}\right)=2r\cos\left(\dfrac{\alpha}{2}\right)}⟶2xsin(2α)=2rcos(2α)\\</p><p>\longrightarrow\sf{x=\dfrac{r}{\tan\left(\dfrac{\alpha}{2}\right)}\quad\quad\dots(4)}⟶x=tan(2α)r…(4)\\</p><p>Similarly, on considering triangles GNF and GBF, since the quadrilateral GNFB is also cyclic, we get,\\</p><p>\longrightarrow\sf{y=\dfrac{r}{\tan\left(\dfrac{\beta}{2}\right)}\quad\quad\dots(5)}⟶y=tan(2β)r…(5)\\</p><p>From ΔABC, we have,\\</p><p>\sf{\cos\alpha=\dfrac{3}{5}}cosα=53\\</p><p>\sf{\cos\beta=\dfrac{4}{5}}cosβ=54\\</p><p>We know that,\\</p><p>\longrightarrow\sf{\tan\left(\dfrac{\theta}{2}\right)=\sqrt{\dfrac{1-\cos\theta}{1+\cos\theta}}}⟶tan(2θ)=1+cosθ1−cosθ\\</p><p>Hence,\\</p><p>\longrightarrow\sf{\tan\left(\dfrac{\alpha}{2}\right)=\sqrt{\dfrac{1-\cos\alpha}{1+\cos\alpha}}}⟶tan(2α)=1+cosα1−cosα\\</p><p>\longrightarrow\sf{\tan\left(\dfrac{\alpha}{2}\right)=\sqrt{\dfrac{1-\dfrac{3}{5}}{1+\dfrac{3}{5}}}}⟶tan(2α)=1+531−53\\</p><p>\longrightarrow\sf{\tan\left(\dfrac{\alpha}{2}\right)=\dfrac{1}{2}}⟶tan(2α)=21\\</p><p>And,</p><p>\longrightarrow\sf{\tan\left(\dfrac{\beta}{2}\right)=\sqrt{\dfrac{1-\cos\beta}{1+\cos\beta}}}⟶tan(2β)=1+cosβ1−cosβ\\</p><p>\longrightarrow\sf{\tan\left(\dfrac{\beta}{2}\right)=\sqrt{\dfrac{1-\dfrac{4}{5}}{1+\dfrac{4}{5}}}}⟶tan(2β)=1+541−54\\</p><p>\longrightarrow\sf{\tan\left(\dfrac{\beta}{2}\right)=\dfrac{1}{3}}⟶tan(2β)=31\\</p><p>Hence (4) becomes,\\\longrightarrow\sf{x=\dfrac{r}{\left(\dfrac{1}{2}\right)}}⟶x=(21)r\\\longrightarrow\sf{x=2r}⟶x=2r\\And (5) becomes,\\\longrightarrow\sf{y=\dfrac{r}{\left(\dfrac{1}{3}\right)}}⟶\\</p><p></p><p>$